$\frac{x^2+3x-40}{\left(x-5\right)^2}$

$\left(x^3+8x^2+11x-20\right)-\left(3x+x^2-4\right)$

$\frac{1-3x^5}{x^3+3}$

$\int\left(8x^3cos^2\left(x^4\right)\right)dx$

$\sin\left(180+a\right)=-\sin\left(a\right)$

$\left(-1\:+\:2j\right)^2$

$\left(-x^4+5x^2-3x-1\right)-\left(5x^4+3-x^2-2x+3\right)$